Theorem abeq1 2886

Description: Equality of a class variable and a class abstraction. Commuted form of abeq2 2885. (Contributed by NM, 20-Aug-1993.)

Assertion

Ref	Expression
abeq1	⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem abeq1

Step	Hyp	Ref	Expression
1		abeq2 2885	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
2		eqcom 2766	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
3		bicom 225	. . 3 ⊢ ((𝜑 ↔ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ↔ 𝜑))
4	3	albii 1822	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
5	1, 2, 4	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Syntax hints:   ↔ wb 209  ∀wal 1537   = wceq 1539   ∈ wcel 2112  {cab 2736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2159  ax-12 2176  ax-ext 2730

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1542  df-ex 1783  df-nf 1787  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831

This theorem is referenced by:  rabeqc  3603  disjOLD  4349  dm0rn0  5772  dffo3  6866  dfsup2  8955  rankf  9270  fmla0xp  32875  dfon3  33779  dfiota3  33810  scottabf  41367  dffo3f  42222